# Prove that the dot product map between a vector space and its dual is a linear isomorphism

Let us define a map $$\phi:\mathbb R^n\rightarrow{\mathbb R^n}^*, \phi(v)=L_v$$ where $$L_v:\mathbb R^n\rightarrow{\mathbb R}, w\rightarrow v\cdot w$$

Now prove $$\phi$$ is a linear isomorphism.

As far as linearity goes, $$\phi(v+w)=L_{v+w}$$.

Now, $$L_{v+w}(p)=(v+w)\cdot p=v\cdot p+ w\cdot p=L_v(p)+L_w(p)$$ for all $$p\in\mathbb{R}$$. Thus $$\phi(v+w)=L_{v+w}=L_v+L_w$$. Similarly it is easy to prove scalar multiplication. Thus, $$\phi$$ is a linear map.

But I cannot prove isomorphism. My guess is to use the fact that for a linear map, injectivity (which can be proved by the fact that only the zero vector maps to the zero) can be used to proved isomorphism. Can someone show how to prove isomorphism?

Showing that for finite dimensional spaces $$V$$ we have $$\dim V=\dim V^*$$ is quite easy, and it is independent of your claim, so let's use this fact. We want to prove injectivity of $$\phi$$, and this, together with the fact that $$\phi$$ maps one vector space to another one with the same dimension, will yield the claim.
Now, suppose $$\phi(v)=0 \in \mathbb{R}^{n*}$$, so that $$L_v(w)=(v,w)=0$$ for every $$w$$. So also for $$w=v$$, so that $$(v,v)=|v|^2=$$, and thus $$v=0$$.
Suppose that $$v \in \mathbb R^n$$ is a vector that $$\phi$$ maps to zero, that is, $$L_v : \mathbb R^n \to \mathbb R$$ is the zero function. Thus, we want to show that $$v=0$$. Well, observe that $$0 = L_v(v) = v {\,\small \bullet\,} v = |v|^2$$ and this implies that $$|v|=0$$, which (of course) only happens when $$v=0$$. Hence, $$\phi$$ is injective, and since $$\dim ((\mathbb R^n)^*) = \dim(\mathbb R^n)$$, it follows that it is surjective also.
Since $$dim\mathbb{R}^n=dimHom(\mathbb{R}^n,\mathbb{R})$$, it suffices to verify injectivity of your map. Suppose $$\phi(v)=0$$. So, $$L_{v}=0$$ . Hence for all $$w\in \mathbb{R}^n$$, $$L_v(w)=0=v\cdot w=0$$. Put $$w=v$$. So, $$||v||=0$$ which implies that $$v=0$$.
Recall: $$T:\mathbb{R}^n\rightarrow \mathbb{R}^m$$ is injective if and only if $$kerT=\{0\}$$